Contour integral for a sine-squared trigonometric quotient ($J(a,b)=2\pi (a-\sqrt{a^2-b^2})/b^2$)

For $a>b>0$, substitution $z=e^{i\theta }$ gives poles at zero and the two roots of $b{z}^2+2az+b$. The inside-pole residues sum to $(a-\sqrt{a^2-b^2})/(i{b}^2)$, soThe double pole at zero is essential; omitting it gives the wrong sign and value.